Chapter 13 Hydrocarbons Solutions
Question - 11 : - What are the necessary conditions for any system to be aromatic?
Answer - 11 : -
The necessary conditions for amolecule to be aromatic are:
· It should have a single cyclic cloudof delocalised n-electrons above and below the plane of the molecule.
· It should be planar. This is becausecomplete delocalization of n-electrons is possible only if the ring is planarto allow cyclic overlap of p-orbitals.
· It should contain Huckel number ofelectrons, i.e., (4n + 2) n-electrons where n = 0, 1, 2, 3 etc.
A molecule which does not satisfy any one or more of the above conditions issaid to be non-aromatic.
Question - 12 : - Explain why the following systems are not aromatic?
Answer - 12 : -
Due to the presence of asp3-hybridized carbon, the system is not
planar. It does contain six n-electrons but the system is not fully conjugatedsince all the six n-electrons do not form a single cyclic electron cloud whichsurrounds all the atoms of the ring. Therefore, it is not an aromatic compound.
(ii)
Due to the presence of a sp3 -carbon, the system is not planar. Further,itcontains only four n-electrons, therefore, the system is not aromatic becauseit does not contain planar cyclic cloud having (4n + 2) n-electrons.
(iii)
Cyclo-octatetraene is not planar but is tub shaped. It is, therefore, anon-planar system having 8 n-electrons.
Therefore, the molecule is not aromatic since it does not contain a planarcyclic cloud having (4n + 2) n-electrons.
Answer - 13 : - (i)The two substituents in the benzene ring are present at p-positions. Therefore,the sequence of reactions should be such that first an o, p-directing group,i.e., Br atom should be introduced in the benzene ring and this should befollowed by nitration. Thus,
(ii) Here since the two substituents are at p-position w.r.t. each other,therefore, the first substituent in the benzene ring should be a o, p-directinggroup (i.e., CH3) and then the other group (i.e., NO2)should be introduced. Therefore, the sequence of reactions is:
(iii)Here since the two substituents are at m-position w.r.t. each other,therefore, the first substituent in the benzene ring should be a m-directinggroup (i.e., NO2) and then other group (i.e.,Cl) should beintroduced.
(iv)Acetophenone can be prepared by F.C. acylation using either acetyl chlorideor acetic anhydride.
Question - 14 : - In the alkane, CH3CH2—C(CH3)2—CH2—CH(CH3)2,identify 1°, 2°, 3° carbon atoms and give the number of H-atoms bonded to eachone of these.
Answer - 14 : -
The expanded formula of the givencompound is
Question - 15 : - What effect does branching of an alkane chain has on its boiling point?
Answer - 15 : - Branchingof carbon atom chain decreases the boiling point of alkane.
Question - 16 : - Addition of HBr to propene yields 2-bromopropane, while in presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
Answer - 16 : - Additionof HBr to propene is an ionic electrophilic addition reaction in which theelectrophile, i.e., H+ first adds to give a more stable 2°carbocation. In the 2nd step, the carbocation is rapidly attacked by thenucleophile Br~ ion to give 2-bromopropane.
In presence of benzoyl peroxide, the reaction is still electrophilic but theelectrophile here is a Br free radical which is obtained by the action ofbenzoyl peroxide on HBr
In the first step, Br radical adds to propene in such a way so as to generatethe more stable 2° free radical. In the second step, the free radical thusobtained rapidly abstracts a hydrogen atom from HBr to give 1-bromopropane.
From the above discussion, it is evident that although both reactions areelectrophilic addition reactions but it is due to different order of additionof H and Br atoms which gives different products.
Question - 17 : - Write down the products ofozonolysis ofl,2-dimethylbenzene (o-xylene). How does the result support Kekule structure ofbenzene?
Answer - 17 : -
o-Xylene may be regarded as aresonance hybrid of the following two Kekule structures. Ozonolysis of each oneof these gives two products as shown below:
Thus, in all, three products are formed. Since all the three products cannot beobtained from any one of the two Kekule structures, this’ shows that o-xyleneis a resonance hybrid of the two Kekule structures (I and II).
Question - 18 : - Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.
Answer - 18 : - Thehybridization state of carbon in these three compounds is:
Since s-electrons are closer to the nucleus, therefore, as the s-character ofthe orbital making the C—H bond increases, the electrons of C—H bond lie closerand closer to the carbon atom. In other words, the partial +ve charge on theH-atom and hence the acidic character increases as the s-character of the orbitalincreases. Thus, the acidic character decreases in the order: Ethyne >Benzene > Hexane
Question - 19 : - Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
Answer - 19 : -
Due to the presence of an electroncloud containing 6 n-electrons above and below the plane of the ring, benzeneis a rich source of electrons. Consequently, it attracts the electrophiles(electron-deficient) reagents towards it and repels nucleophiles (electron-rich) reagents. As a result, benzene undergoes electrophilic substitutionreactions easily and nucleophilic substitutions with difficulty.
Question - 20 : - How will you convert the following compounds into benzene?
(i) Ethyne (ii) Ethene (iii) Hexane.
Answer - 20 : -